AMC 10 · 2013 · #20

Grade 8 geometry-2d
Archetype Compound Area by Decomposition / Difference └ Sub-archetype Circle-Polygon Hybrid
area-circlesrotation-isometrycircular-sectorsymmetry-argument identify-subproblemssymmetry-argument ↑ Prerequisites: area-circles
📏 Long solution 💡 3 insights
Problem
A unit square is turned 45° about its center. As it turns, its inside passes over a region of the plane. Find the area of that swept region.

Pick an answer.

(A)
$1 - \frac{\sqrt2}{2} + \frac{\pi}{4}$
(B)
$\frac{1}{2} + \frac{\pi}{4}$
(C)
$2 - \sqrt2 + \frac{\pi}{4}$
(D)
$\frac{\sqrt2}{2} + \frac{\pi}{4}$
(E)
$1 + \frac{\sqrt2}{4} + \frac{\pi}{8}$

AMC 10 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

The swept blob has a curvy outline and no single formula for it, so Tool #7 (Identify Subproblems) is the engine: cut the region into pieces whose areas we already know how to find. The cut is guided by Tool #17 (Visualize Spatial Relationships) — picturing how each corner traces a circular arc while the sides stay closer in — and Tool #14 (Extreme Principle), which says the farthest any point travels is the corner distance, so that distance is the outer radius. The whole picture is unchanged by a 90° turn, and that symmetry lets us solve just two representative pieces (one round, one flat) and multiply.

1STEP 1

Put it on axes, find the reach

Center at the origin, corners at (±1/2,±1/2): a corner reaches out √2/2, a side only 1/2, so corners run the rim.

OC=√((1/2)²+(1/2)²)=√2/2, O-to-side=1/2
2STEP 2

Slice into eight wedges

Draw a ray every 45°. By quarter-turn symmetry the eight wedges come in only two kinds: four arc wedges, four flat wedges.

area=4×(one arc wedge)+4×(one flat wedge)
3STEP 3

The four arc wedges make a half-disk

Each arc wedge is a 45° slice of the corner's circle, so four of them make a half-disk of radius √2/2, area π/4.

4×(45° slice)=1/2π(√2/2)²=π/4
4STEP 4

One flat wedge by coordinates

Shoelace on the first-octant wedge O,(√2/2,0),(1/2,√2/2-1/2),(1/2,1/2) gives (2-√2)/4 each, so four flat wedges total 2-√2.

[OP₁QP₂]=1/2|√2/2 (√2/2-1/2)+1/2·1/2-1/2 (√2/2-1/2)|=(2-√2)/4, ×4=2-√2
5STEP 5

Add the round part and the flat part

Round part plus flat part: π/4+(2-√2)=2-√2+π/4≈1.37, which is choice (C).

π/4+(2-√2)=2-√2+π/4≈1.37 → (C)
Answer
2 - √2 + π/4
The answer must beat the square's own area, since the starting square (0°) is part of the sweep: indeed 1.37 ＞ 1. It must also stay below the disk of radius √2/2 that contains everything, whose area is π(√2/2)²=π/2≈1.57: indeed 1.37 ＜ 1.57. So 1 ＜ 2-√2+π/4 ＜ π/2, exactly where a 45° sweep should land. The nearby distractors miss: (B) ≈1.29 is too small (its flat part 1/2 undercounts 2-√2≈0.59) and (D) ≈1.49 is too big, so only (C) fits both bounds.
💡Key takeaway

Spin the square and split the trail into eight wedges: four rounded ones fold into a half-circle of radius √2/2 (π/4) and four straight ones add up to 2-√2, for a total of 2-√2+π/4, choice (C).

  • Put it on axes, find the reach
  • Slice into eight wedges
  • The four arc wedges make a half-disk
  • One flat wedge by coordinates
  • Add the round part and the flat part

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